Introduction to the Prisoners Versus Guards Game

INTRODUCTION TO THE PRISONERS VERSUS GUARDS GAME

arXiv:0801.0833v1 [math.CO] 6 Jan 2008

TIMOTHY HOWARD, EUGEN J. IONASCU, AND DAVID WOOLBRIGHT

1. Introduction Suppose that you are competing in a two-player game in which you and your opponent attempt to pack as many “prisoners” as possible on the squares of an n × n checkerboard; each prisoner has to be “protected” by an appropriate number of guards. Initially, the board is covered entirely with guards. The players – designated as “red” and “blue” – take turns adjusting the board configuration using one of the following rules in each turn: I. Replace one guard with a prisoner of the player’s color. II. Replace one prisoner of either color with a guard and replace two other guards with prisoners of the player’s color. That is, each player takes a turn increasing the total number of prisoners by one. We require that, at every stage of the game, each prisoner lies adjacent to at least as many guards as the number of the other prisoners adjacent it. The squares adjacent to a given square are those squares, situated directly above, below, to the left, to the right, or diagonal to the square in question. An arrangement of prisoners and guards that satisfies this requirement and has exactly one occupant per square is called a valid board. The game ends when neither player can further adjust the board using rules I and II while maintaining a valid board. The player whose color represents more prisoners is the winner. This is the game of Prisoners and Guards – a game that can be played and analyzed without extensive knowledge of mathematics. Figure 1 depicts students from Blackmon Road Middle School, in Columbus, Georgia, trying their hands at the game. We invite the reader to play the game online by running the Java Applet available at http://csc.colstate.edu/woolbright/. The guards in this game are related to the half domination set in the king’s graph as introduced in in a paper by Hoffman, Laskar, and Markus (see [2]). Similar domination problems are studied have been studied by Bode, Harborth, and Harporth (see [1]) and by Watkins, Ricci, and McVeigh (see [6]). The Prisoners and Guards game originated as a puzzle created by the third author with a focus on minimizing the size of the dominating set (the guards). Date: January 21, 2007. Key words and phrases. upper bounds, winning strategies. 1

2


Figure 1. Blackmon Road Middle School Students Play PvG

In the two-player game, one fundamental question that naturally arises is “How do we decide when the game is over?” The short answer is that the game is over when the board configuration has reached a maximal state. A valid board to which no adjustments can be made to increase the total number of prisoners is called a maximal board. One can also define a maximum configuration as being a maximal arrangement of prisoners and guards that has the greatest number of prisoners of all valid boards. For n ∈ {1, 2, 3} all maximal boards are also maximum configurations. We will see example 4 × 4 boards that are maximal but do not contain maximum configurations. Let P (n) denote the number of prisoners in a maximum configuration. Characterizing the sequence {P (n)}∞ n=1 will help us determine when to end the game. It also proves an interesting avenue for exploration on its own. Since the lone square on a 1 × 1 board has no adjacent squares, we can place a prisoner in it and be sure that there are at least as many guards as prisoners lying in all adjacent squares – none. Therefore, we have P (1) = 1. By exhaustively checking all sixteen 2 × 2 cases, we find eleven valid boards, each having zero, one, or two prisoners. Thus, P (2) = 2. We analyze the cases n = 3, 4, 5, and 6 in sections 2 and 3. Exact values for P (n), n ∈ {7, 8, 9, 10, 11}, can be found in a paper by Ionascu, Pritikin, and Wright (see [5]), who employ linear programming techniques in the study of P (n). In Section 4 we also obtain an upper bound on P (n); this is about the best that we can say for n ≥ 12.


3

2. The game analysis for n = 3 and n = 4 Playing Prisoners and Guards on a 1 × 1 board or on a 2 × 2 board is not all that interesting. When we increase the board size slightly and consider the game on a 3 × 3 board, strategy becomes more of a factor. We will see that the arrangement in Figure 2 is a maximum configuration, as we establish in Theorem 2.2. We use the diamond to represent prisoners and blank squares represent guards.

.

Figure 2. Maximum 3x3 Board In fact, this is the only maximal arrangement (up to a rotation). Let us observe that a maximal board permits no adjustments using either Rule I or Rule II. First we consider arrangements that are maximal with respect to Rule I (i.e. one cannot simply add more prisoners in the existing configuration). Perhaps it is not difficult to convince oneself that any valid board having zero, one, or two prisoners can be adjusted using Rule I; after factoring out rotations and reflections, there are ten unique cases to check. Therefore, a maximal board must have at least three prisoners. Figure 3 depicts all valid boards (up to rotations and reflections) that contain three, four, or five prisoners and are maximal with respect to Rule I.

Figure 3. 3 × 3 Boards that are Maximal w.r.t. Rule I Each one of these configurations can be adjusted to match the arrangement in Figure 2 by using Rule II (and one or more adjustments using Rules I and II in some of the cases). It follows that a maximal 3 × 3 board must contain at least six prisoners. We record this fact in the following lemma. LEMMA 2.1. Any maximal 3 × 3 board has at least six prisoners. With Figure 2, we see that a valid 3 × 3 configuration can have six prisoners. Does there exist a valid configuration with more than six prisoners? Suppose that a 3 × 3 board arrangement contains seven prisoners (and two guards). Since there are four non-corner edge squares, a prisoner must occupy at least one of them. Since this prisoner lies adjacent to at most two guards, the

4


board cannot be valid. These observations, Lemma 2.1, and the fact that Figure 2 depicts a valid configuration with six prisoners lead us to the following conclusion. THEOREM 2.2. A maximum 3 × 3 board contains six prisoners, i.e. P (3) = 6. As a matter of fact, the configuration shown in Figure 2 is the only maximal 3 × 3 board (up to a rotation of the board). From this we learn that the second player has a good chance to win by using Rule II all of the time. The first player may force a tie if she can lead the board configuration in such a manner that will require her opponent to use Rule I. In fact, this is manageable if she plays into the pattern in Figure 2, forcing the blue to use Rule I in the last step and so the number of prisoners of each color will end up equal. We now turn our attention to 4 × 4 boards. This board size proves interesting because there exist many maximal arrangements that are not maximum configurations. We include some known maximal arrangements with eight prisoners in Figure 4, but there may be others.

Figure 4. Some Maximal 4 × 4 Boards If we factor out rotations and reflections of the board, there are three maximum arrangements as depicted in Figure 5. We obtained these via an exhaustive search of all 4 × 4 valid boards and verified that there are no other equivalence classes.

,

,

.

Figure 5. Maximum 4 × 4 Boards To prove something about maximum 4 × 4 board configurations, it helps to dissect the board and consider what can happen in the vicinity of the corner squares. Suppose that we have a 2 × 2 block C of squares situated in one corner of the board. If the corner square within C does not contain a guard, then it contains a prisoner. If the latter is the case, then C must contain at least two guards. Thus we have established the following fact.


5

LEMMA 2.3. If C is a 2 × 2 corner block within a valid board (n ≥ 2), then it must contain at least one guard. With this in mind, we are equipped to consider maximum 4 × 4 boards by partitioning them into four 2 × 2 corner blocks and following through with the consequences. This will lead us to the conclusion of the next proposition. PROPOSITION 2.4. P (4) = 9. That is, every maximum 4 × 4 valid board has nine prisoners. PROOF. Since the configurations in Figure 5 are valid and each contains nine prisoners, it follows that P (4) ≥ 9. It is enough to show that P (4) ≤ 9. We assume that there exist a valid board B with ten or more prisoners. By dropping prisoners if necessary, we can say there are exactly ten. We shall see that this leads to a contradiction. We partition B into four 2 × 2 blocks as indicated in Figure 6(a).

, (a)

G G (b)

G G

,

.

(c)

Figure 6. Block Partitions of a 4 × 4 Board

Since by assumption the board contains ten prisoners, it follows from Lemma 2.3 that at least two of these blocks must contain three prisoners each. Without loss of generality, assume that the upper left block is one of them. We see in the proof of Lemma 2.3 that the lone guard must lie in square b11 , as depicted in Figure 6(b). For the board to be valid, the non-corner edge prisoners in b12 and b21 must each lie adjacent to three guards. This is only possible if guards lie in the squares b13 , b23 , b31 , and b32 , as Figure 6(b) indicates. As previously noted, at least two of the blocks must contain three prisoners each. The only way to achieve this will be for the lower right block to have three prisoners, with a guard in b44 as shown in Figure 6(c). Now we see that the prisoners situated in squares b34 and b43 necessitate the presence of guards in squares b24 and b42 . By placing prisoners in all squares not yet committed, we will have a total of only eight prisoners on the board, contradicting our assumption that the board has ten prisoners. Thus, our assumption was invalid.

6


3. Analysis of the 5 × 5 and 6 × 6 Cases In our analysis of 5 × 5 and 6 × 6 board configurations, we will partition the boards into 3 × 3 blocks. The following lemma will help in the examination of these blocks. LEMMA 3.1. If C is a 3 × 3 corner block within a valid board (n > 3), then it must contain at least three guards. Moreover, if C contains exactly three guards, then it must contain a prisoner diagonally opposite (within C) to the corner square. PROOF. Assume that there exists a valid n × n (n > 3) board configuration with a 3 × 3 corner block C that contains only two guards. Without loss of generality, suppose that C is situated in the upper left corner of the board. Since four guards are required to protect a prisoner residing on an interior square, and we have two guards, a guard must occupy c22 . Since, by assumption, there remains only one more guard, there must lie a prisoner in c12 or c21 . However, three guards are required to cover a prisoner that is situated in a non-corner edge square. Hence, the board cannot be valid and we have a contradiction. To establish the last part of the claim, observe that besides the guard at c22 the other two are either at c12 or c21 , or they must lie adjacent to the prisoner located at c12 or c21 . (If one of the guards were located in c33 then we would have a prisoner in either c12 or c21 without a sufficient number of adjacent guards). Possible arrangements, up to a reflection about the main diagonal, appear in Figure 7. In these cases, we are left with a prisoner in c33 .

Figure 7. Possible 3 × 3 UL Corner Blocks With 3 Guards Now, we are ready to consider the 5×5 case. The only maximum configuration (up to rotations) is illustrated in Figure 8. Proposition 3.2 establishes that this is a maximum 5×5 board configuration.

Figure 8. The Maximum 5 × 5 Board Configuration

PROPOSITION 3.2. A maximum 5 × 5 board configuration contains fifteen prisoners; that is, P (5) = 15.


7

PROOF. Assume that there exists a valid 5 × 5 board configuration with 16 or more prisoners. We will see that this leads to a contradiction. Divide the 5 × 5 board into two opposite (overlapping) corner 3 × 3 blocks, A and C, that have a square in common and two 2 × 2 opposite corner blocks, B and D, as illustrated in Figure 9(a).

A A A D D

A A B B A A B B A A/C C C D C C C D C C C

A A A D D

(a)

A A A A/C D C C C D C C C

A A A A

A/C C C C C

(b)

(c)

Figure 9. Partitions of the 5 × 5 Board According to Lemma 3.1, the two 3 × 3 blocks A and C collectively contain at most 2(6) − 1 = 11 prisoners since the shared square (common to blocks A and C) must contain a prisoner. Recall that Lemma 2.3 establishes that each of blocks B and D contains at most three prisoners. Thus, for the board to contain a total of sixteen prisoners we must find either ten or eleven prisoners shared in blocks A and C. (They cannot share just nine prisoners since that would force a 2 × 2 block to hold four prisoners). Case 1. Blocks A and C share 11 prisoners. In this case, one of the 2 × 2 blocks holds three prisoners and the other holds two prisoners. Without loss of generality, let us suppose that the B block has three prisoners; then the one guard must lie in the (1,5) position. To cover the three prisoners in block B, guards in the (1,3), (2,3), (3,4), and (3,5) positions (refer to Figure 9(b)). For the board to be valid, block A must match one of the corner blocks depicted in Figure 7, none of which allows guards in both the (1,3) and the (2,3) positions. Therefore, the board is not valid and we have a contradiction in this case. Case 2. Blocks A and C share 10 prisoners. In this case, each of the 2 × 2 blocks B and D holds three prisoners. In order to maintain a valid board configuration, we are then forced to place guards in the (1,3), (1,5), (2,3), (3,1), (3,2), (3,4), (3,5), (4,3), and (5,3) positions as indicated in Figure 9(c). But then there remain only nine uncommitted squares in which to place the ten prisoners that blocks A and C are supposed to share. Thus, we also find a contradiction in this case. For n = 6 all maximum boards amount to rotations or small perturbations of the arrangement in Figure 10, the validity of which yields the lower bound P (6) ≥ 22. We will show that, in fact, P (6) = 22 by an analysis of manageable size. Dunbar, Hoffman, Laskar, and Markus assert

8


(without proof) a fact about 1/2-domination in the king’s graph dimension 6 which, if true, implies that P (6) = 22 (see [2]). This is indeed the case, as we shall establish next. We use a more specific version of Lemma 3.1 in order to obtain this fact.

Figure 10. A Maximum 6 × 6 Board Configuration

LEMMA 3.3. If C is a 3×3 corner block holding six prisoners within a valid board (n > 3), where c11 is the corner square, then up to a diagonal symmetry the block has one of the six arrangements in Figure 7. PROOF. The position c22 must have a guard as we have seen and also one of the positions c12 or c21 must have a guard. By symmetry we can assume we have a guard at c12 . Then there are six possible spots for the third guard. This gives exactly the six arrangements in Figure 7. PROPOSITION 3.4. A maximum 6 × 6 board contains twenty two prisoners. That is, P (6) = 22. PROOF. We have observed that P (6) ≥ 22. To verify that P (6) ≤ 22 let us assume the existence of a valid arrangement C with 23 prisoners; we shall see that this leads to a contradiction. By Lemma 3.1, three of the four 3 × 3 corner blocks have six prisoners and one has five prisoners. Without loss of generality, we may assume that the block with five prisoners is the lower right one. By Lemma 3.3 and by symmetry, we can assume that the block in the upper left corner is one of those in Figure 7 and the upper right 3 × 3 corner block is an arrangement found in Figure 11. Since one of these arrangements has diagonal symmetry we have only seven possible situations as listed Figure 11.

Figure 11. Possible Upper Right 3 × 3 Corner Blocks In each case one can find that the prisoner at c13 or c24 does not have enough guards around it. This contradicts the existence of a configuration with 23 prisoners.


9

We suspect that this block partition approach can be adapted to compute or bound P (n) for larger sizes of n, although this approach could turn out to be quite lengthy. These proofs may very well be pursued as undergraduate research projects. 4. Upper bound for P (n) and the deficiency function As the board size grows larger, establishing the exact number of prisoners on a maximum board becomes increasingly difficult. The proof of Proposition 2.4 foreshadows the importance of finding useful upper bounds on P (n). In this section, we construct a tool that will help in establishing these bounds – the deficiency matrix. We then use the deficiency matrix to determine a general upper bound for P (n). Suppose that we have fixed the board size at n×n (n ≥ 3). With each configuration, we associate a binary matrix X = (xij ) defined by

1 if a prisoner lies in the (i, j) position 0 if a guard lies in the (i, j) position. Many who work in combinatorics and graph theory, such as Hedetniemi, Hedetniemi, and Reynolds xij =

(see [4]) have employed this idea. In any local measure of optimality we must be attentive to the number of prisoners lying in square adjacent to a particular square; we let x∗ij denote the number of prisoners lying in squares adjacent to the (i, j) square. The deficiency matrix serves as an ad-hoc, local measure of the optimality of a given board configuration. Its construction arises from our observations and conjectures of maximum board configurations. We define the deficiency matrix δ = (δij ) by δij = expectation − x∗ij , where  1 if (i, j) is a corner square with xij = 1     2 if (i, j) is a corner square with xij = 0    2 if (i, j) is an edge square with xij = 1 expectation = 4 if (i, j) is an edge square with xij = 0     4 if (i, j) is an interior square with xij = 1    6 if (i, j) is an interior square with x = 0. ij Figure 12 depicts a 4 × 4 non-maximal board configuration and its corresponding deficiency

matrix. The positive entries in the deficiency matrix indicate areas of the board that are thought

to be less than optimal; the 2’s indicate that the “worst” deficiencies occur on the corresponding interior squares. Since we use these values to obtain an upper bound on P (n), it helps to first consider bounds on δij for 1 ≤ i, j ≤ n. Suppose that the (i, j) square contains a prisoner. If (i, j) is a corner square, then at least two of the three adjacent squares must contain guards; therefore x∗ij ≤ 1 and

10


0 1 0 0

1 0 2 0

0 2 0 1

0 0 1 0

Figure 12. Non-maximal 4 × 4 Board and Its Deficiency Matrix

in checking our expectation value above we see that δij ≥ 0. Likewise, if (i, j) is an edge square containing a prisoner or an interior square with a prisoner, we find that δij ≥ 0. Suppose that we find a guard in a corner square (i, j). Then three squares lie adjacent to this square, so we find at most three prisoners in the neighboring squares. Therefore, x∗ij ≤ 3 and so δij ≥ −1. Via similar considerations, we find that if a guard occupies an edge square then δij ≥ −1 and for an interior square we get δij ≥ −2. We define the net deficiency of a board configuration as the sum of all entries in the deficiency matrix,

∆=

n X

δij .

i,j=1

For instance, the board configuration in Figure 12 has a net deficiency of 8. We expect maximum board configurations to correspond to minimum net deficiency values. We will relate ∆ to the overall number of guards in a given board configuration. Let PC and GC denote the total number of prisoners and guards, respectively, found in the corner squares. Similarly, PE and GE refer to the prisoners and guards in edge squares, and PI and GI refer to prisoners and guards in interior squares. With this notation we have

P P P ∆ = corners δij + edges δij + interior δij ≥ (0 · PC − 1 · GC ) + (0 · PE − 1 · GE ) + (0 · PI − 2 · GI ) = −GC − GE − 2 · GI .

This establishes the next lemma.

LEMMA 4.1. The net deficiency of a given configuration satisfies the inequality ∆ ≥ −GC − GE − 2 · GI .


11

Now we are ready to think about bounding the size of P (n). Observe that ( 8 − δij , if xij = 1 and (i, j) is an interior square 4xij + x∗ij = 6 − δij , if xij = 0 and (i, j) is an interior square (1)

3xij + x∗ij =

(

5 − δij , if xij = 1 and (i, j) is an edge square 4 − δij , if xij = 0 and (i, j) is an edge square

2xij + x∗ij =

(

3 − δij , if xij = 1 and (i, j) is a corner square 2 − δij , if xij = 0 and (i, j) is a corner square.

THEOREM 4.2. The number of prisoners in a valid configuration is given by (2)

P = PROOF.

1 3n2 4n − + (3PE + 6PC − ∆). 5 5 10

Summing the left hand sides of the equations in (1) over all squares of the board,

we obtain 4·PI +3·PE +2·PC +

X

x∗ij = 4·PI +3·PE +2·PC +8·PI +5·PE +3·PC = 12·PI +8·PE +5·PC .

1≤i,j≤n

We will equate this result with the sum of the right hand sides. In summing over the interior squares that contain prisoners, this contributes 8 − δij = 6 + 2 − δij for each such square , whereas the interior square that contain guards contribute only 6−δij per square. There are (n−2)2 interior P squares, so altogether these sum to 6(n − 2)2 + 2 · PI − int. sqrs. δij . Similarly summing the right P hand sides over all edge squares we get 4 · 4 [4(n − 2)] + 1 · PE − edge sqrs. δij . Summing over the P corners yields 8 + 1 · PC − corner sqrs. δij . Combining these right-hand sums and equating with the left-hand sum, we obtain the equation

12PI + 8PE + 5PC = 6(n − 2)2 + 2PI + 16(n − 2) + PE + 8 + PC − ∆ or 10PI + 7PE + 4PC = 6n2 − 8n − ∆. Then since P = PI + PE + PC we then obtain 10P = 6n2 − 8n + 3PE + 6PC − ∆, which leads to (2). By combining the inequality in Lemma 4.1 with this theorem, we obtain a crude upper bound on P (n).

12


COROLLARY 4.3. In a maximum configuration of prisoners and guards on a n × n board the number of prisoners obeys the inequality (3)

P (n) ≤

2n2 + n . 3

PROOF. Using Lemma 4.1 and (2) we obtain P (n) ≤ = = This implies

3n2 4n 1 − + (3PE + 6PC + 2GI + GE + GC ) 5 5 10 i 3n2 4n 1h 2PE + 5PC + 2(n − 2)2 − 2PI + 4(n − 2) + 4 − + 5 5 10 i 3n2 4n 1h 2PE + 5PC + 2(n − 2)2 − 2P + 2PE + 2PC + 4n − 4 . − + 5 5 10

1 3n2 4n − + (4PE + 7PC + 2n2 − 4n + 4) 5 5 10 i 2 4n 1h 3n (4)(4)(n − 2) + (7)(4) + 2n2 − 4n + 4 − + ≤ 5 5 10 2 4n + 2n = 5 2 5 4n + 2n 2n2 + n Therefore P (n) ≤ . = 6 5 3 We believe that ∆ ≤ O(n) in general. This fact is equivalent to P (n) ≤ 3n2 /5 + O(n). However 1 1+ 5

P (n) ≤

we can tighten the upper bound (3) by getting a better estimate for ∆. LEMMA 4.4. In a valid configuration the net deficiency satisfies ∆ ≥ −1 · G = −1(GI + GE + GC ). PROOF. Recall our previous observations about the possible range of values for δij . If the (i, j) board position contains a prisoner then from the definition it follows that δij ≥ 0. If the square is a corner or edge square containing a guard then δij ≥ −1. For an interior square containing a guard, we have noted that δij ≥ −2. Let us focus on this last case. Suppose that a guard occupies the (i, j) interior position in a valid board configuration and that δij = −2. Then it must be the case that all adjacent squares contain prisoners, as depicted in Figure 13(a). The g’s denote guards that are then forced into the arrangement in order for the configuration to be valid. We see that each of the prisoners in the squares diagonally adjacent to this position lies adjacent to five or six guards (depending on the occupants in the squares marked with asterisks). The possible deficiency values for neighboring squares appear in Figure 13(b). Summing these deficiency values, we find that the net contribution of the 3 × 3 block satisfies 2 ≤ ∆local ≤ 6. Notice that, as the g’s in Figure 13(a) suggest, it is not possible for two such 3 × 3


13

blocks around guards with deficiency −2 to overlap. Thus, we see that each guard on the board contributes a net deficiency value not less than −1. ∗ g g g ∗

g g

g g g g

∗ g g g ∗

0 1, 2 -2 0 1, 2 0 1, 2 1, 2

0

(a)

(b)

Figure 13. Local Configuration Near a Guard with Deficiency -2 Summing the δij ’s over all board positions, we have ∆ =

P

prisoners δij

+

P

guards δij

≥ −1 · G.

Using this bound on ∆ in Theorem 4.2, we obtain a better upper bound for P (n). The calculations parallel those used in the proof of Corollary 4.3. THEOREM 4.5. For an n × n maximum arrangement of prisoners and guards, the number of prisoners, P (n), satisfies the inequality (4)

P (n) ≤

7n2 + 4n . 11

PROOF. By Lemma 4.4, −∆ ≤ G. Applying this upper bound in Theorem 4.2, we get i 3 i 3 2 4 1h 4 1h 3PE + 6PC + G = n2 − n + 3P + 3PC − 3PI + G P ≤ n − n+ 5 5 10 5 5 10 i 1h 3 2 4 2P + 3 PC − PI + n2 . n − n+ = 5 5 10 2 Subtracting 10 P from both sides and combining the n2 terms, we see that this implies 8 7 2 4 3 P ≤ n − n+ PC − PI 10 10 5 10 3 7 2 4 n − n+ 4 − P + PE + PC ≤ 10 5 10 7 2 4 3 ≤ n − n+ 4 − P + 4n − 4 . 10 5 10 The claim now follows after a bit of arithmetic. It seems that this method of finding an upper bound can be further sharpened, but we have not found a complete analysis to show that ∆ ≥ −O(n). This would be the sharpest result for this type of domination problem. 5. Other Results, Conjectures and Open Questions Ionascu, Pritikin, and Wright have established values of P (n) for n ∈ {7, 8, 9, 10, 11} (see [5]). Most of their arrangements were obtained using the LPSolve IDE program in the Lesser GNU public

14


domain for solving integer linear programming problems with Branch-and-Bound and Simplex Methods. The second author used CPLEX while visiting at the Georgia Institute of Technology in the Faculty Development Program in 2005-2006; with the help of Professor William Cook he analyzed the case n = 11. Many interesting questions remain to be answered. What are the values of P (n) for integers n larger than 11? With error-free play, does one particular player enjoy an advantage? Perhaps the advantage varies with the board size. If P (n) is odd, we conjecture that the game favors the red player, but it is not clear that a winning strategy exists. When P (n) is even, we suspect that error-free play by both players will lead to a tie. Given that we find several maximal 4 × 4 board configurations with eight prisoners (an even number), it seems that the second player (blue) will find opportunities to win unless s/he is forced to use Rule I. The question is: can the red player always achieve a win or a tie? We believe there is a strategy for the red player to win despite all of these chances for the blue player. In general, it is apparent that that the final maximal configuration is an important factor in the game, since the number of prisoners in it determines the fate of the game. So it is in the red player’s interest to end in a maximal arrangement with an odd number of prisoners on the board. Similarly it is part of blue strategy to divert the end configuration to a maximal one that has an even number of prisoners. Each player can change the configuration at only one place at which the opponent has already directed the game toward his final configuration and leave one place as it is. As a result, almost half the prisoners on the final board configuration are where each player wanted them to be. So from this perspective the end game is dictated by the parity and the number of maximal configurations with P (n) − 1, P (n) − 2, ... prisoners. 7n2 + 4n , but we conjecture that 11 2 P (n) = 3n /5 + O(n). It would be great to see someone, especially a student, improve our upper In this paper, we have shown that P (n) is bounded above by

bound. Acknowledgments. Special thanks go to Annalisa Crannell, of Franklin and Marshall College, for her many helpful comments and suggestions for this paper. We also thank Mike McCoy, a Columbus State Univeresity student who spent hours designing a program that has a very good run-time, even for large size boards, and is producing arrangements that we think are “very close” to maximum arrangements.

References [1] J-P. Bode, H. Harborth, M. Harporth, King independence on triangle boards, Discrete Mathematics, 266 (2003), pp. 101-107.


15

[2] J.E. Dunbar, D.G. Hoffman, R.C. Laskar, L.R. Markus, α- Domination , Discrete Mathematics, 211 (2000), pp. 11-26. [3] O. Favaron, On a conjecture of Fink and Jacobson concerning k-domination and k-dependence, J. Combin. Theory Ser. B 39 (1985), no. 1, pp. 101-112. [4] S.M. Hedetniemi, S.T. Hedetniemi and R. Reynolds, Combinatorial problems on chessboards: II chapter 6 (pp. 133-62) in: T.W. Haynes, S.T. Hedetniemi, P.J. Slater, (Eds.), Domination in Graphs: Advanced Topics, Marcel Dekker, New York (1998). [5] E. J. Ionascu, D. Pritikin, S. E. Wright, k-Dependence and Domination in Kings Graphs, accepted by the American Mathematical Monthly to appear in 2008. [6] J. Watkins, C. Ricci, B. McVeigh, King’s domination and independence: a tale of two chessboards, Congresus Numerantium, 158(2002), pp.59-66. Current address: Math Department, Columbus State University, 4225 University Avenue, Columbus, GA 31907 E-mail address: [email protected]; E-mail address: ionascu [email protected]; E-mail address: woolbright [email protected];